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Normal form theorem for graphs of groups. Let be a graph of groups and .

Any can be written as
as before.

If , this expression includes `backtracking’, meaning that for some , with , and furthermore that if , then .

Similar to the case of the free group, the proof boils down to the fact that the Bass–Serre tree is a tree.

Proof. To simplify notation, set , so

.

Fix base points in the vertex spaces , which are chosen to coincide when the vertices do. Then is a loop in based at , and is a path, crossing the corresponding edge space, from to . This allows us to consider as a loop in based at . (We may assume by adding letters from a maximal tree.)

Consider the universal covering and fix a base point over in . Let be the lift of based at and its image in the Bass–Serre tree . We now analyze and closely.

Choose adjoining and so that the edge traversed by when lifted at corresponds to the coset .

Then lifts to a path in which terminates at . Similarly, lifts at to a path across the edge to the vertex space terminating at . Therefore, lifts at to a path which crosses the edge space and ends at .

Then, lifts at to a path in ending at , and lifts at to a path across the edge into the vertex space , and terminating at . Thus lifts at to a path which crosses , through , across , and ending at

.

We continue this process until we have explicitly constructed . By hypothesis, , so and are both loops in and , respectively. Since is a tree, must backtrack.

This implies that and that . That is, by Lemma 18,

.

Therefore, we have found a backtracking, and can accordingly shorten . This proves the theorem.

Let be a group and with . Then we call residually finite (hereafter RF) if there exists a subgroup of finite index such that . In other words, for every nontrivial element of , there exists a finite index subgroup that does not contain that particular element.

Example. Finite groups are RF, since the trivial subgroup has finite index and does not contain any of the nontrivial elements of .

Metaquestion. How general is the class of RF groups? In particular, which finitely generated/finitely presented groups are RF?

Remarks. Assume that is finitely generated.

(i) The definition can be strengthened so that we may assume that our finite index subgroup is normal in . Indeed, if is finitely generated, then there are only finitely many subgroups of a given fixed index . (See the exercise from the second lecture.) If , and is a subgroup of finite index that does not contain , then

is a subgroup of finite index in which excludes , since and have the same index in for all , so this intersection is really the intersection of finitely many subgroups of finite index. Thus also has finite index in .

(ii) Equivalently, is RF if and only if for each not the identity, there exists a homomorphism , where is a finite group, for which is not the identity in . Indeed, if is RF and is normal subgroup provided by (i), then does not die under the natural homomorphism from to . Conversely, given such a homomorphism, the kernel of is a subgroup of finite index that does not contain .

(iii) Also, is RF if and only if

That is, the intersection of all the subgroups of finite index in is the trivial subgroup. Were some nonidentity element to be contained in this intersection, then it would be contained in each subgroup of finite index in , so this element prevents from being RF. Conversely, if this intersection is trivial, each nonidentity element of must be excluded from some finite index subgroup, so is RF.

(iv) If is RF and , then there exists a finite index subgroup with for all . Here, just take the intersection of the finite index subgroup associated with each . This is again a finite index subgroup of .

Lemma 2: Let be a finitely generated group.

(i) If is RF and is a subgroup, then is RF.

(ii) If is RF and with finite index in , then is RF.

That is, RF passes to all subgroups and also to supergroups of finite index.

Proof. For (i), choose . Considered as an element of , there exists a homomorphism to a finite group for which is not the identity. The restriction of to is also a homomorphism to a finite group for which the image of is nontrivial. The kernel of this restricted homomorphism is a subgroup of finite index in that does not contain . (Note that we did not assume that is finitely generated.)

For (ii), choose . If , then is a finite index subgroup not containing , and we are done. If , then there exists a finite index subgroup that does not contain . However, is also a finite index subgroup of , so we are done.

Topological reformulation of RF

Now, we would like to connect RF with a topological property of a space with fundamental group . Let be a compact manifold with universal covering and . Accordingly, we assume throughout that is finitely generated. (We will see later that the manifold condition can be relaxed.)

Theorem 2: The group is RF if and only if the following condition holds: for every compact subset there exists a finite sheeted covering for which embeds homeomorphically in .

Proof. Assume the topological condition holds and choose any not the identity. This corresponds to a loop (based at a point dependent only upon our choice of universal covering) in which we also denote by . To this loop, there also is a corresponding lift to a connected arc in with distinct endpoints and .

Let be the compact set . Then, there exists a finite sheeted covering for which and are distinct points of . By the lifting property of covering spaces, this implies that if is the subgroup of finite index corresponding to the covering , then . Therefore is RF.

Conversely, suppose that is RF and choose any subset . Then, since acts freely and properly discontinuous on , the set of those (not the identity) for which intersects nontrivially is finite. Now, choose of finite index containing none of the elements of . If is the corresponding finite sheeted covering, we have that for all . That is, embeds homeomorphically in .

Remark. We never really used here that was a manifold, only that its fundamental group acted properly discontinuous on the universal covering. (The action need not be free either, since this would only add another finite number of elements to our set .) Thus, it suffices to assume that is Hausdorff and locally compact.

Examples.

(1) Finitely generated abelian groups are RF. (Exercise.)

(2) Selberg’s Lemma (Malcev-Selberg): If is a finitely generated linear group, that is, for some , then is RF.

Proof. We begin with the case . Since RF passes to arbitrary subgroups, it suffices to prove that is RF.

Choose any . This means that the matrix has some nonzero entry, say . Since is an integer, we can find some large prime that does not divide . Now, consider the homomorphism

given by reducing the entries of a given matrix modulo . This is a homomorphism because matrix multiplication is linear in the entries. (Exercise – make this precise.) Since is a finite field, its general linear group is a finite group, i.e. the kernel of is a finite index subgroup of , and it does not contain by construction. Thus is RF, and thus any subgroup thereof is also RF.

More generally, if is finitely generated, we can arrange that , where is the subring generated by the entries of a finite collection of generators for . In particular, this is an integral domain, since it is a finitely generated subring of . Thus, given a nonzero element of as above, we can find a prime ideal of that does not divide , i.e. .

Now, is a finite integral domain, that is, a finite field, and we can build a reduction homomorphism analogous to the situation over the integers. The the image of the reduction homomorphism is the general linear group of a finite field, so it is a finite group. We now proceed exactly as above.

This theorem is often referred to as Selberg’s Lemma, even though Malcev supposedly proved it first.

(3) Free groups of finite rank are linear, so they are RF. Notice that we have seen, in the exercise from the first lecture, that is a subgroup of for all , so it suffices to prove that is linear. One such representation the subgroup of generated by the two matrices

There are several ways to prove that this group is free of rank two. First, one use a bit of hyperbolic geometry and the action of on the hyperbolic plane to prove that this group is the fundamental group of a hyperbolic manifold that is deformation equivalent to the rose with two petals. Also, one can use the so-called Ping-Pong Lemma, which says that a finite collection of homeomorphisms of a space that satisfy a certain set of conditions necessarily generate a free subgroup of the homeomorphism group.